How do you solve $2cos^2(x/5)-3cos(x/5)+1=0$ ?

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Answer 1 · 2015-08-04T04:35:29

Solve 2cos^2 (x/5) - 3cos (x/5) + 1 = 0
Ans: 0, $10pi.$ $+- (5pi)/3$

Call $cos (x/5) = t$ and solve quadratic equation:

$2t^2 - 3t + 1 = 0$

Since a + b + c = 0, two real roots are: 1 and $c/a = 1/2.$
Next, solve:
a. $t = cos (x/5) = 1$ -->

$x/5 = 0$ , --> $x = 0$

and $x/5 = 2pi$ --> $x = 10pi$

b. $t = cos (x/5) = 1/2$ --> $x/5 = +- pi/3$ --> $x = +- (5pi)/3$

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